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Math Help - Partial Fractions

  1. #1
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    Partial Fractions

    I'm having trouble setting up the partial fractions for this integral

    \int{\frac{x^2-3x+7}{(x^2-4x+6)^2}}

    I think this is the case of a repeating, irreducible quadratic factor. So, would the decomposition be this?

    \frac{A}{(x^2-4x+6)} + \frac{B}{(x^2-4x+6)^2}
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  2. #2
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    Quote Originally Posted by centenial View Post
    I'm having trouble setting up the partial fractions for this integral

    \int{\frac{x^2-3x+7}{(x^2-4x+6)^2}}

    I think this is the case of a repeating, irreducible quadratic factor. So, would the decomposition be this?

    \frac{A}{(x^2-4x+6)} + \frac{B}{(x^2-4x+6)^2}
    Note quite remember when you have a quadratic in the denominator the numerator have the form

    Ax+B so it should look like this

    \displaystyle \frac{Ax+B}{(x^2-4x+6)} + \frac{Cx+D}{(x^2-4x+6)^2}
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  3. #3
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    Thanks so much!
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  4. #4
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    Just in case you also want to do it by 'the method of avoiding partial fractions':

    Observe that (x^2-3x+7)-(x^2-4x+6)-(x+1) = 0. Then:

    Spoiler:
    \displaystyle \int{\frac{x^2-3x+7}{(x^2-4x+6)^2}}\;{dx} =  \int{\frac{(x^2-3x+7)-(x^2-4x+6)-(x+1)+(x^2-3x+7)}{(x^2-4x+6)^2}}\;{dx}
    Spoiler:

    \displaystyle =  \int{\frac{(x^2-3x+7)-(x^2-4x+6)}{(x^2-4x+6)^2}}\;{dx}-\int{\frac{(x+1)-(x^2-3x+7)}{(x^2-4x+6)^2}}\;{dx}

    Spoiler:
    \displaystyle =  \int{\frac{x^2-3x+7-x^2+4x-6}{(x^2-4x+6)^2}}\;{dx}-\int{\frac{x+1-x^2+3x-7}{(x^2-4x+6)^2}}\;{dx}

    Spoiler:
    \displaystyle =  \int{\frac{x+1}{(x^2-4x+6)^2}}\;{dx}-\int{\frac{-(x^2-4x+6)}{(x^2-4x+6)^2}}\;{dx}

    Spoiler:
    \displaystyle =  \int{\frac{x+1}{(x^2-4x+6)^2}}\;{dx}+\int{\frac{1}{(x^2-4x+6)}}\;{dx}
    Spoiler:
    You can do your usual integration (completing the square etc) now.

    Last edited by TheCoffeeMachine; December 7th 2010 at 09:12 AM.
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  5. #5
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    Quote Originally Posted by centenial View Post
    I'm having trouble setting up the partial fractions for this integral

    \int{\frac{x^2-3x+7}{(x^2-4x+6)^2}}

    I think this is the case of a repeating, irreducible quadratic factor. So, would the decomposition be this?

    \frac{A}{(x^2-4x+6)} + \frac{B}{(x^2-4x+6)^2}
    Or....denominator is x^2-4x+6 squared.

    If we add an x and a 1, we will have the numerator...

    \displaystyle\frac{\left(x^2-4x+6\right)+(x+1)}{\left(x^2-4x+6\right)^2}=\frac{1}{x^2-4x+6}+\frac{x+1}{\left(x^2-4x+6\right)^2}
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  6. #6
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    Quote Originally Posted by Archie Meade View Post
    Or....denominator is x^2-4x+6 squared.

    If we add an x and a 1, we will have the numerator...

    \displaystyle\frac{\left(x^2-4x+6\right)+(x+1)}{\left(x^2-4x+6\right)^2}=\frac{1}{x^2-4x+6}+\frac{x+1}{\left(x^2-4x+6\right)^2}
    Looks like a zero term made me do works!
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